How do you balance chemical word equations?

I have this chemistry worksheet and I'm quite stumped by it. I've never been that great when it came to science or math, so this is all over the head. Can someone explain to me how to balance these equations?





Directions: Write the word equations below as chemical equations and balance.





1) Aluminum bromide and chlorine gas react to form aluminum chloride and bromine gas.








2) Copper and sulfuric acid react to form copper (II) sulfate and water and sulfur dioxide.








Any help would be greatly appreciated!


Thanks!|||Like a baseball game, it helps to have a scorecard. In other words, what do the words mean when it comes to describing compounds. Once you have surmounted this barrier, balancing follows.





Start with "aluminum bromide". If you look it up, it is AlBr3. Why? Aluminum can shed 3 outer electrons to form an Al+3 ion and Br (which, as an element, exists as a dimer, Br-Br), can "suck up" one electron per atom to form Br-. Chlorine is also a dimer (Cl-Cl). However, Cl is more adapt at "sucking up" electrons than is Br, so it can boot the Br- out of the compound, take its electron, and leave a Br atom (which rapidly reacts with another unhappy Br that has been booted out). This may sound silly, but it is about what happens, and leads us to balance. The conversion to compounds and dimers leads to....


AlBr3 + Cl2 -%26gt; AlCl3 + Br2


Balancing an equation simply means that every atom on one side is accounted for on the other. Here, the chlorines and bromines are out of balance. So how do we even them out?? Well, if we double the AlBr3 (and we have to double the whole compound, and not just one part), we get 6 bromines on the left. Since the bromine winds up in a dimerized form, for those 6 bromines, they can be balanced by 3Br2. The same deal with Chlorine. So then, we have


2 AlBr3 + 3 Cl2 -%26gt; 2 AlCl3 + 3Br2.





On to #2. The sulfuric acid (H2SO4) serves to oxidize elemental copper to the +2 state. Some of the acid provides SO4-2 ion for the copper salt. However, if something goes "up" in oxidation state, something else goes "down". And that something is the "S" in H2SO4, which drops from +6 oxidation to +4 oxidation in SO2.


The rules of oxidation states is pretty easy. Elements are 0, hydrogen in a compound is +1 and oxygen is -2. Compounds have to be overall 0. There are exceptions, but not too many to worry about at this stage.





In this mess, some additional water is formed, since we have to do something with the 4Os in each molecule of H2SO4; 2 wind up in SO2, the other 2 wind up in water. So lets try to balance this


Cu + H2SO4 -%26gt; CuSO4 + SO2 + H2O


For starters, we need more H2SO4 per atom of Cu. Lets double it to get...


Cu + 2H2SO4 -%26gt; CuSO4 + SO2 + H2O.


Looks better,but doubling the H2SO4 means we have 4H on the right and 2 on the left. Also, we have 8O on the left and seven on the right.


So lets double the amount of water formed.....


Cu+ 2H2SO4 -%26gt; CuSO4 + SO2 + 2H2O


Hopes this helps.|||Step 1. Write the chemical equation. --- To do this, you need to know what the words mean.





Step 2. Balance the chemical equation. --- The number of atoms on the left has to be the same as on the right. i.e. if the are 2 Cl on the left, there needs to be 2 Cl on the right.





Step 3 (or maybe step 0.5) Open your text book and read it. --- To do this, pick up you chemistry text and put it in front of you. Grab the cover and open the book. I am guessing that this material is somewhere in the first five chapters. You might want to start at the beginning and go from there. While reading, think about what you are reading, reread what you don't understand. Follow through the worked examples, and maybe try some of the sample problems. Ask your teacher about any questions that you may have. They are called a teacher for a reason, because they teach you. There is no excuse for sucking your way through life, especially when science and math are so essential to everything.|||1) Cl2 + 2AlBr3 --%26gt; 2AlCl + 3Br2


2) Cu + 2H2SO4 --%26gt; CuSO4 + 2H2O + SO2|||1) 2AlBr3 + 3Cl2 --%26gt; 2AlCl3 + 3Br2


2) Cu + 2H2SO4 --%26gt; CuSO4 + 2H2O + SO2